Lab Manual

1. 1
Estimation of iron in Mohr’s salt (Ferrous ammonium sulphate)
solution by permanganometry titration
Dr. Mausumi Adhya
HOD and Associate Professor
Supreme Knowledge Foundation, West Bengal, India

2. 2
Theory
Mohr’s salt is a double salt of ferrous sulphate and ammonium sulphate and the formula is FeSO4.
(NH4)2SO4. 6H2O. In Mohr’s salt iron is in +2 oxidation state and can be estimated by using potassium
permanganate (KMnO4) in acid medium. Hence the titration is called permanganometry titration.
The reaction between KMnO4 and Mohr’s salt occurs depending on reduction potential value.
MnO4
- + 8H+ + 5e = Mn2+ + 4H2O Eo= 1.51 V
Fe3+ + e= Fe2+ Eo= 0.77 V
I.e. reduction potential value of MnO4
-/ Mn2+ is more than Fe3+/ Fe2+. Therefore, MnO4
- is reduced to
Mn2+ and Fe2+ is oxidized to Fe3+. KMnO4 is a strong oxidizing agent in presence of acid.
Overall reaction
Reduction MnO4
- + 8H+ + 5e = Mn2+ + 4H2O
Oxidation 5Fe2+ - 5e= 5Fe3+
MnO4
-+ 8H++5Fe2+= Mn2++ 5Fe3++4H2O
KMnO4 is a secondary standard and oxalic acid is used to find out the strength of KMnO4.
MnO4
- + 8H+ + 5e = Mn2+ + 4H2O Eo= 1.51 V
2CO2 +2e = C2O4
2- Eo=-0.49V
Here reduction potential value of MnO4
-/ Mn2+ is more than 2CO2 / C2O4
2-. Hence MnO4
- is reduced
to Mn2+ and C2O4
2- is oxidized to 2CO2.
Overall reaction
Reduction 2 x (MnO4
- + 8H+ + 5e = Mn2 ++ 4H2O)
Oxidation 5x (C2O4
2- = 2CO2 +2e)
2 MnO4
- +16H+ + 5C2O4
2- = 2Mn2+ +10CO2 + 4H2
In this titration no redox indicator is used because pink coloured KMnO4 acts as a self indicator.
Materials
1. Apparatus: Burette with burette stand, pipette, conical flask, beaker, volumetric flask, reagent
bottle, measuring cylinder, dropper
2. Chemicals: Mohr salt, oxalic acid, KMnO4, 2 (N) H2SO4, H3PO4
Procedure
1. Preparation of reagents
(I) Preparation of 250 ml 0.1 (N) oxalic acid solution
Molecular weight of oxalic acid (H2C2O4.2H2O) = 2+24+64+36=126
Gram equivalent weight of oxalic acid =126/2 g=63 g

3. 3
To prepare 250 ml 0.1 (N) solution gm equivalent of oxalic acid required= [63×250]/ [1000×10] =1.575
g
1.575 g of oxalic acid is accurately weight out in 250 ml volumetric flask and dissolved in distilled
water with continuous stirring. The water is added upto the mark. Finally the solution is made
uniform by shaking.
(II) Preparation of 1000 ml 0.1 (N) KMnO4 solution
In acid medium the reaction is
MnO4- + 8H+ + 5e = Mn+2 + 4H2O
Molecular weight of KMnO4= 39+55+64=158
Equivalent weight of KMnO4 = Molecular weight of KMnO4/number of electrons gained per
molecule = 158/5 g= 31.6 g
To prepare 1000 ml 0.1 (N) KMnO4 solution gm equivalent required=31.6/10 g =3.16 g
About 3.16 gm KMnO4 is taken in a 500 ml beaker. The solid is dissolved in about 300-400 ml distilled
water by stirring with a glass rod. The solution is filtered in a 1000 ml volumetric flask through glass
funnel containing glass wool. The beaker, glass wool and funnel are washed well with remaining
distilled water. The solution is shaken for making the solution uniform. The clear filtered solution of
KMnO4 is kept in a dark brown colour glass bottle.
(III) Preparation of 1000 ml 2(N) H2SO4
Strength of concentrated H2SO4 is 36N
V1S1=V2S2
V× 36= 1000× 2 i.e. V=55.56 ml
55.56 ml concentrated H2SO4 is added slowly with constant stirring by a glass rod in 800-900 ml
distilled water containing in a 1000 ml measuring flask. Distilled water is added upto the mark. The
solution is shaken for making the solution uniform.
(IV) Preparation of 250 ml 0.1 (N) Mohr’s salt solution
Gram equivalent weight of Mohr’s salt = molecular weight of Mohr salt/1= 392.13 gm
To prepare 250 ml 0.1 (N) Mohr’s salt gm equivalent required=[392.13×250]/[1000×10]= 9.803 g
9.8 g of Mohr’s salt is weight out in 250 ml volumetric flask and dissolved in 2(N) H2SO4 with
continuous stirring. 2(N) H2SO4 is added upto the mark. Finally the solution is made uniform by
shaking.
2. Standardization of KMnO4 by standard oxalic acid
10 ml oxalic acid is pipetted out in a 250 ml conical flask. 60 ml 2(N) H2SO4 is added to it. The mixture
is heated on an asbestos board to about 70-80ºC. Therefore, the solution is titrated in hot condition

4. 4
with KMnO4 running from burette dropwise. Disappearance of pink colour at the first time is slow at
the beginning then becomes rapid as Mn+2 which is formed in the reaction, autocatalysis the reaction.
The end point is faint pink colour of solution by a drop of KMnO4. Experiment is repeated three
times.
3. Estimation of iron in Mohr’s salt
10 ml Mohr’s salt is pipetted out in a 250 ml conical flask. 50 ml 2 (N) H2SO4 and 2 ml H3PO4 are
added to it. The mixture is titrated with KMnO4 until a faint pink colour persists in the solution.
Experiment is repeated three times.
Results
Table 1: Standardization of KMnO4 by standard oxalic acid
Number of
observation
Volume of
oxalic acid
(ml)
Burette reading Mean volume
of KMnO4
(ml)
Strength of
oxalic acid (
N)
Strength of
KMnO4
(N)
Initial Final Actual
1
10 V 0.1 S
2
3
Strength of KMnO4 = [10×0.1]/V = S (N)
Table 2: Estimation of iron in Mohr’s salt
Number of
observation
Volume
of
Mohr’s
salt (ml)
Burette reading Mean
volume of
KMnO4 (ml)
Strength
of
KMnO4
(N)
Amount of Fe2+
in Mohr salt
(g/L)
Initial Final Actual
1
10 X S
2
3
Calculation
MnO4
- + 8H++5Fe2+= Mn2 ++ 5Fe3++4H2O
Here, 5 equivalents MnO4
- reacts 10 equivalents Fe2+ i.e. 1 equivalent MnO4
- reacts 2 equivalent Fe2+.
1000 ml 1(N) KMnO4 = 55.85 gm Fe+2
X ml S (N) KMnO4= (55.85×X×S)/1000 g Fe+2
Now 10 ml Mohr’s salts solution contains (55.85×X×S)/1000 g Fe+2
Therefore, 1000 ml Mohr’s salts solution contains= (55.85×X×S×1000)/(1000×10) g Fe+2=
(55.85×X×S)/10 g Fe+2
The amount of iron present in Mohr’s salt solution
= [55.85× Total volume of KMnO4 required for titration of Mohr’s salt solution × strength of KMnO4 / volume of Mohr’s
salts solution] g /L

5. 5
Conclusion
The amount of iron present in Mohr’s salt solution is ……….g/L
Precautions
1. During titration of oxalic acid by KMnO4, the temperature should not fall below 70ºC.
2. Mohr salt should not be dissolved in distilled water as the salt is hydrolyzed. 2 (N) H2SO4 is
used as a solvent.
3. KMnO4 should be kept in coloured bottle as KMnO4 decomposes in bright light.
Viva Voce
1. What is the formula of Mohr’s salt?
2. What is the oxidation state of iron Mohr’s salt?
3. What is the formula of oxalic acid?
Answer. (COOH)2. 2H2O
4. Find out the gm-equivalent weight of potassium permanganate in acid medium.
5. Find out the gm-equivalent weight of oxalic acid.
6. Why KMnO4 is a secondary standard solution?
Answer. KMnO4 is a secondary standard as in crystalline KMnO4 contains MnO2 as impurity and
again aqueous solution of KMnO4 decomposes in presence of bright light.
4MnO4
-+2H2O= 4MnO2 + 4OH- + 3O2
7. How can you find out the exact strength of KMnO4?
8. How can you determine the amount of iron present in Mohr’s salt. Discuss in details with
reactions.
9. Why KMnO4 acts as a strong oxidant in acid medium and oxidizes Fe+2 of Mohr’s salt to Fe+3
and oxalic acid to CO2?
10. Why H3PO4 is used for titration of Mohr’s salt by KMnO4?
H3PO4 masks the yellow colour of Fe+3 by forming [Fe (HPO4)] + and helps in the sharp end point.
11. Why no indicator is required for estimation of iron in Mohr’s salt by KMnO4?
Answer. In this titration no indicator is used because pink coloured KMnO4 acts as a self indicator.